Find the first two non vanishing maclaurin terms
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Find the first two nonvanishing terms in the Maclaurin series of $sin(x + x^3)$.
Suggestion: use the Maclaurin series of $sin(y)$ and write $y = x + x^3$
Using
this result, find
$limlimits_{xto 0}frac{sin(x + x^3)−x}{x^3}$
$sin(y)= y-frac{y^3}{3!}+frac{y^5}{5!}+frac{y^7}{7!}$
$y= x+x^3$
$x+x^3-frac{(x+x^3)^3}{3!}+frac{(x+x^3)^5}{5!}-frac{(x+x^3)^7}{7!}$
Now what do I do?
limits taylor-expansion
add a comment |
up vote
3
down vote
favorite
Find the first two nonvanishing terms in the Maclaurin series of $sin(x + x^3)$.
Suggestion: use the Maclaurin series of $sin(y)$ and write $y = x + x^3$
Using
this result, find
$limlimits_{xto 0}frac{sin(x + x^3)−x}{x^3}$
$sin(y)= y-frac{y^3}{3!}+frac{y^5}{5!}+frac{y^7}{7!}$
$y= x+x^3$
$x+x^3-frac{(x+x^3)^3}{3!}+frac{(x+x^3)^5}{5!}-frac{(x+x^3)^7}{7!}$
Now what do I do?
limits taylor-expansion
Please make sure I did not change the context of your post.
– homegrown
Dec 9 '14 at 19:52
no you didn't thank you
– Tk706
Dec 9 '14 at 19:58
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Find the first two nonvanishing terms in the Maclaurin series of $sin(x + x^3)$.
Suggestion: use the Maclaurin series of $sin(y)$ and write $y = x + x^3$
Using
this result, find
$limlimits_{xto 0}frac{sin(x + x^3)−x}{x^3}$
$sin(y)= y-frac{y^3}{3!}+frac{y^5}{5!}+frac{y^7}{7!}$
$y= x+x^3$
$x+x^3-frac{(x+x^3)^3}{3!}+frac{(x+x^3)^5}{5!}-frac{(x+x^3)^7}{7!}$
Now what do I do?
limits taylor-expansion
Find the first two nonvanishing terms in the Maclaurin series of $sin(x + x^3)$.
Suggestion: use the Maclaurin series of $sin(y)$ and write $y = x + x^3$
Using
this result, find
$limlimits_{xto 0}frac{sin(x + x^3)−x}{x^3}$
$sin(y)= y-frac{y^3}{3!}+frac{y^5}{5!}+frac{y^7}{7!}$
$y= x+x^3$
$x+x^3-frac{(x+x^3)^3}{3!}+frac{(x+x^3)^5}{5!}-frac{(x+x^3)^7}{7!}$
Now what do I do?
limits taylor-expansion
limits taylor-expansion
edited Dec 30 '14 at 16:19
Antonio Vargas
20.6k244111
20.6k244111
asked Dec 9 '14 at 19:27
Tk706
116211
116211
Please make sure I did not change the context of your post.
– homegrown
Dec 9 '14 at 19:52
no you didn't thank you
– Tk706
Dec 9 '14 at 19:58
add a comment |
Please make sure I did not change the context of your post.
– homegrown
Dec 9 '14 at 19:52
no you didn't thank you
– Tk706
Dec 9 '14 at 19:58
Please make sure I did not change the context of your post.
– homegrown
Dec 9 '14 at 19:52
Please make sure I did not change the context of your post.
– homegrown
Dec 9 '14 at 19:52
no you didn't thank you
– Tk706
Dec 9 '14 at 19:58
no you didn't thank you
– Tk706
Dec 9 '14 at 19:58
add a comment |
2 Answers
2
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oldest
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up vote
0
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Note that the formula for Maclaurin Series is
$$f(x)=sum_{n=0}^{infty}frac{f^{(n)}(0)}{n!}x^n.$$
So calculating the first few terms with $y=x+x^3$ gives us
$$sin(y)=frac{0}{0!}y^0+frac{1}{1!}y^1-frac{0}{2!}y^2-frac{1}{3!}y^3+frac{0}{4!}y^4+...\=y-frac{y^3}{3!}+frac{y^5}{5!}-...$$
Now, if we plug $x+x^3$ back into $sin(y),$ we need to make sure that we have the correct derivatives. So, we have for $f(x)=sin(x+x^3)$
$$f^{(1)}(0)=(1+3x^2)cos(x+x^3)=1\f^{(3)}(0)=6cos(x+x^3)-18x(1+3x^2)(sin(x+x^3))-(3x^2+1)(cos(x+x^3))=6-1=5$$since you only need the first two non vanishing terms. Now we have
$$sin(x+x^3)=x-frac{5x^3}{6}.$$
Now we need to find $$limlimits_{xto 0}frac{sin(x+x^3)-x}{x^3}.$$Plugging the series in for $sin(x+x^3)$ we have
$$limlimits_{xto 0}frac{x-frac{5x^3}{6}-x}{x^3}=limlimits_{xto 0}frac{frac{5x^3}{6}}{x^3}=limlimits_{xto 0}frac{5x^3}{6x^3}=frac56.$$
Hope this helps.
add a comment |
up vote
0
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jnh's answer is correct, other than a sign error in the middle of it which is corrected by the end.
But you can do this without taking any derivatives by hand. The trick is to partially expand the binomial. You have:
$$sin(x+x^3)=sum_{n=0}^infty frac{(x+x^3)^{2n+1} (-1)^n}{(2n+1)!}$$
Considering the binomial theorem, the first term contributes a term of order $1$ and a term of order $3$, while the second term contributes a term of order $3$ and higher order terms. More generally the $n$th term contributes a term of order $2n+1$ and higher order terms, so going up to $n=1$ retrieves the first two terms of the desired series.
The first two terms in the overall series will be order $1$ and $3$. The term of order $1$ can just be read off; it is just $x$. The term of order $3$ is a combination of the contribution from the $n=0$ term ($x^3$) and the contribution from the $n=1$ term, which by the binomial theorem is $frac{-x^3}{6}$.
So we have $sin(x+x^3)=x+frac{5x^3}{6}+o left ( x^3 right )$. (The $o left ( x^3 right )$ is actually of order $5$, but this is not required to solve the problem.) Plugging this in we get
$$frac{sin(x+x^3)-x}{x^3}=frac{frac{5x^3}{6}+o left ( x^3 right )}{x^3} = frac{5}{6} + o(1)$$
and so the desired limit is $frac{5}{6}$. This approach of combining terms of the same order is common in perturbation theory problems in physics.
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Note that the formula for Maclaurin Series is
$$f(x)=sum_{n=0}^{infty}frac{f^{(n)}(0)}{n!}x^n.$$
So calculating the first few terms with $y=x+x^3$ gives us
$$sin(y)=frac{0}{0!}y^0+frac{1}{1!}y^1-frac{0}{2!}y^2-frac{1}{3!}y^3+frac{0}{4!}y^4+...\=y-frac{y^3}{3!}+frac{y^5}{5!}-...$$
Now, if we plug $x+x^3$ back into $sin(y),$ we need to make sure that we have the correct derivatives. So, we have for $f(x)=sin(x+x^3)$
$$f^{(1)}(0)=(1+3x^2)cos(x+x^3)=1\f^{(3)}(0)=6cos(x+x^3)-18x(1+3x^2)(sin(x+x^3))-(3x^2+1)(cos(x+x^3))=6-1=5$$since you only need the first two non vanishing terms. Now we have
$$sin(x+x^3)=x-frac{5x^3}{6}.$$
Now we need to find $$limlimits_{xto 0}frac{sin(x+x^3)-x}{x^3}.$$Plugging the series in for $sin(x+x^3)$ we have
$$limlimits_{xto 0}frac{x-frac{5x^3}{6}-x}{x^3}=limlimits_{xto 0}frac{frac{5x^3}{6}}{x^3}=limlimits_{xto 0}frac{5x^3}{6x^3}=frac56.$$
Hope this helps.
add a comment |
up vote
0
down vote
Note that the formula for Maclaurin Series is
$$f(x)=sum_{n=0}^{infty}frac{f^{(n)}(0)}{n!}x^n.$$
So calculating the first few terms with $y=x+x^3$ gives us
$$sin(y)=frac{0}{0!}y^0+frac{1}{1!}y^1-frac{0}{2!}y^2-frac{1}{3!}y^3+frac{0}{4!}y^4+...\=y-frac{y^3}{3!}+frac{y^5}{5!}-...$$
Now, if we plug $x+x^3$ back into $sin(y),$ we need to make sure that we have the correct derivatives. So, we have for $f(x)=sin(x+x^3)$
$$f^{(1)}(0)=(1+3x^2)cos(x+x^3)=1\f^{(3)}(0)=6cos(x+x^3)-18x(1+3x^2)(sin(x+x^3))-(3x^2+1)(cos(x+x^3))=6-1=5$$since you only need the first two non vanishing terms. Now we have
$$sin(x+x^3)=x-frac{5x^3}{6}.$$
Now we need to find $$limlimits_{xto 0}frac{sin(x+x^3)-x}{x^3}.$$Plugging the series in for $sin(x+x^3)$ we have
$$limlimits_{xto 0}frac{x-frac{5x^3}{6}-x}{x^3}=limlimits_{xto 0}frac{frac{5x^3}{6}}{x^3}=limlimits_{xto 0}frac{5x^3}{6x^3}=frac56.$$
Hope this helps.
add a comment |
up vote
0
down vote
up vote
0
down vote
Note that the formula for Maclaurin Series is
$$f(x)=sum_{n=0}^{infty}frac{f^{(n)}(0)}{n!}x^n.$$
So calculating the first few terms with $y=x+x^3$ gives us
$$sin(y)=frac{0}{0!}y^0+frac{1}{1!}y^1-frac{0}{2!}y^2-frac{1}{3!}y^3+frac{0}{4!}y^4+...\=y-frac{y^3}{3!}+frac{y^5}{5!}-...$$
Now, if we plug $x+x^3$ back into $sin(y),$ we need to make sure that we have the correct derivatives. So, we have for $f(x)=sin(x+x^3)$
$$f^{(1)}(0)=(1+3x^2)cos(x+x^3)=1\f^{(3)}(0)=6cos(x+x^3)-18x(1+3x^2)(sin(x+x^3))-(3x^2+1)(cos(x+x^3))=6-1=5$$since you only need the first two non vanishing terms. Now we have
$$sin(x+x^3)=x-frac{5x^3}{6}.$$
Now we need to find $$limlimits_{xto 0}frac{sin(x+x^3)-x}{x^3}.$$Plugging the series in for $sin(x+x^3)$ we have
$$limlimits_{xto 0}frac{x-frac{5x^3}{6}-x}{x^3}=limlimits_{xto 0}frac{frac{5x^3}{6}}{x^3}=limlimits_{xto 0}frac{5x^3}{6x^3}=frac56.$$
Hope this helps.
Note that the formula for Maclaurin Series is
$$f(x)=sum_{n=0}^{infty}frac{f^{(n)}(0)}{n!}x^n.$$
So calculating the first few terms with $y=x+x^3$ gives us
$$sin(y)=frac{0}{0!}y^0+frac{1}{1!}y^1-frac{0}{2!}y^2-frac{1}{3!}y^3+frac{0}{4!}y^4+...\=y-frac{y^3}{3!}+frac{y^5}{5!}-...$$
Now, if we plug $x+x^3$ back into $sin(y),$ we need to make sure that we have the correct derivatives. So, we have for $f(x)=sin(x+x^3)$
$$f^{(1)}(0)=(1+3x^2)cos(x+x^3)=1\f^{(3)}(0)=6cos(x+x^3)-18x(1+3x^2)(sin(x+x^3))-(3x^2+1)(cos(x+x^3))=6-1=5$$since you only need the first two non vanishing terms. Now we have
$$sin(x+x^3)=x-frac{5x^3}{6}.$$
Now we need to find $$limlimits_{xto 0}frac{sin(x+x^3)-x}{x^3}.$$Plugging the series in for $sin(x+x^3)$ we have
$$limlimits_{xto 0}frac{x-frac{5x^3}{6}-x}{x^3}=limlimits_{xto 0}frac{frac{5x^3}{6}}{x^3}=limlimits_{xto 0}frac{5x^3}{6x^3}=frac56.$$
Hope this helps.
answered Dec 9 '14 at 20:15
homegrown
3,42031236
3,42031236
add a comment |
add a comment |
up vote
0
down vote
jnh's answer is correct, other than a sign error in the middle of it which is corrected by the end.
But you can do this without taking any derivatives by hand. The trick is to partially expand the binomial. You have:
$$sin(x+x^3)=sum_{n=0}^infty frac{(x+x^3)^{2n+1} (-1)^n}{(2n+1)!}$$
Considering the binomial theorem, the first term contributes a term of order $1$ and a term of order $3$, while the second term contributes a term of order $3$ and higher order terms. More generally the $n$th term contributes a term of order $2n+1$ and higher order terms, so going up to $n=1$ retrieves the first two terms of the desired series.
The first two terms in the overall series will be order $1$ and $3$. The term of order $1$ can just be read off; it is just $x$. The term of order $3$ is a combination of the contribution from the $n=0$ term ($x^3$) and the contribution from the $n=1$ term, which by the binomial theorem is $frac{-x^3}{6}$.
So we have $sin(x+x^3)=x+frac{5x^3}{6}+o left ( x^3 right )$. (The $o left ( x^3 right )$ is actually of order $5$, but this is not required to solve the problem.) Plugging this in we get
$$frac{sin(x+x^3)-x}{x^3}=frac{frac{5x^3}{6}+o left ( x^3 right )}{x^3} = frac{5}{6} + o(1)$$
and so the desired limit is $frac{5}{6}$. This approach of combining terms of the same order is common in perturbation theory problems in physics.
add a comment |
up vote
0
down vote
jnh's answer is correct, other than a sign error in the middle of it which is corrected by the end.
But you can do this without taking any derivatives by hand. The trick is to partially expand the binomial. You have:
$$sin(x+x^3)=sum_{n=0}^infty frac{(x+x^3)^{2n+1} (-1)^n}{(2n+1)!}$$
Considering the binomial theorem, the first term contributes a term of order $1$ and a term of order $3$, while the second term contributes a term of order $3$ and higher order terms. More generally the $n$th term contributes a term of order $2n+1$ and higher order terms, so going up to $n=1$ retrieves the first two terms of the desired series.
The first two terms in the overall series will be order $1$ and $3$. The term of order $1$ can just be read off; it is just $x$. The term of order $3$ is a combination of the contribution from the $n=0$ term ($x^3$) and the contribution from the $n=1$ term, which by the binomial theorem is $frac{-x^3}{6}$.
So we have $sin(x+x^3)=x+frac{5x^3}{6}+o left ( x^3 right )$. (The $o left ( x^3 right )$ is actually of order $5$, but this is not required to solve the problem.) Plugging this in we get
$$frac{sin(x+x^3)-x}{x^3}=frac{frac{5x^3}{6}+o left ( x^3 right )}{x^3} = frac{5}{6} + o(1)$$
and so the desired limit is $frac{5}{6}$. This approach of combining terms of the same order is common in perturbation theory problems in physics.
add a comment |
up vote
0
down vote
up vote
0
down vote
jnh's answer is correct, other than a sign error in the middle of it which is corrected by the end.
But you can do this without taking any derivatives by hand. The trick is to partially expand the binomial. You have:
$$sin(x+x^3)=sum_{n=0}^infty frac{(x+x^3)^{2n+1} (-1)^n}{(2n+1)!}$$
Considering the binomial theorem, the first term contributes a term of order $1$ and a term of order $3$, while the second term contributes a term of order $3$ and higher order terms. More generally the $n$th term contributes a term of order $2n+1$ and higher order terms, so going up to $n=1$ retrieves the first two terms of the desired series.
The first two terms in the overall series will be order $1$ and $3$. The term of order $1$ can just be read off; it is just $x$. The term of order $3$ is a combination of the contribution from the $n=0$ term ($x^3$) and the contribution from the $n=1$ term, which by the binomial theorem is $frac{-x^3}{6}$.
So we have $sin(x+x^3)=x+frac{5x^3}{6}+o left ( x^3 right )$. (The $o left ( x^3 right )$ is actually of order $5$, but this is not required to solve the problem.) Plugging this in we get
$$frac{sin(x+x^3)-x}{x^3}=frac{frac{5x^3}{6}+o left ( x^3 right )}{x^3} = frac{5}{6} + o(1)$$
and so the desired limit is $frac{5}{6}$. This approach of combining terms of the same order is common in perturbation theory problems in physics.
jnh's answer is correct, other than a sign error in the middle of it which is corrected by the end.
But you can do this without taking any derivatives by hand. The trick is to partially expand the binomial. You have:
$$sin(x+x^3)=sum_{n=0}^infty frac{(x+x^3)^{2n+1} (-1)^n}{(2n+1)!}$$
Considering the binomial theorem, the first term contributes a term of order $1$ and a term of order $3$, while the second term contributes a term of order $3$ and higher order terms. More generally the $n$th term contributes a term of order $2n+1$ and higher order terms, so going up to $n=1$ retrieves the first two terms of the desired series.
The first two terms in the overall series will be order $1$ and $3$. The term of order $1$ can just be read off; it is just $x$. The term of order $3$ is a combination of the contribution from the $n=0$ term ($x^3$) and the contribution from the $n=1$ term, which by the binomial theorem is $frac{-x^3}{6}$.
So we have $sin(x+x^3)=x+frac{5x^3}{6}+o left ( x^3 right )$. (The $o left ( x^3 right )$ is actually of order $5$, but this is not required to solve the problem.) Plugging this in we get
$$frac{sin(x+x^3)-x}{x^3}=frac{frac{5x^3}{6}+o left ( x^3 right )}{x^3} = frac{5}{6} + o(1)$$
and so the desired limit is $frac{5}{6}$. This approach of combining terms of the same order is common in perturbation theory problems in physics.
edited Dec 30 '14 at 16:50
answered Dec 30 '14 at 16:35
Ian
66.9k25084
66.9k25084
add a comment |
add a comment |
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Please make sure I did not change the context of your post.
– homegrown
Dec 9 '14 at 19:52
no you didn't thank you
– Tk706
Dec 9 '14 at 19:58